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f'(x)=3x^2+2ax+b f(1)=a+b+4,代入切線方程:5*1+a+b+4-3=0,即a+b=-6 由題意,切線斜率=f’(1)=-5=3+2a+b,即2a+b=-8 由上兩式解得:a=-2,b=-4
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